Description
Some of Farmer John’sNcows (1 ≤N≤ 80,000) are having a bad hair day! Since each cow is self-conscious about her messy hairstyle, FJ wants to count the number of other cows that can see the top of other cows’ heads.
Each cowihas a specified heighthi(1 ≤hi≤ 1,000,000,000) and is standing in a line of cows all facing east (to the right in our diagrams). Therefore, cowican see the tops of the heads of cows in front of her (namely cowsi+1,i+2, and so on), for as long as these cows are strictly shorter than cowi.
Consider this example:
== == – = Cows facing right –>= = == – = = == = = = = =1 2 3 4 5 6
Cow#1 can see the hairstyle of cows #2, 3, 4Cow#2 can see no cow’s hairstyleCow#3 can see the hairstyle of cow #4Cow#4 can see no cow’s hairstyleCow#5 can see the hairstyle of cow 6Cow#6 can see no cows at all!
Letcidenote the number of cows whose hairstyle is visible from cowi; please compute the sum ofc1throughcN.For this example, the desired is answer 3 + 0 + 1 + 0 + 1 + 0 = 5.
Input
Line 1: The number of cows,N.Lines 2..N+1: Linei+1 contains a single integer that is the height of cowi.
Output
Line 1: A single integer that is the sum ofc1throughcN.
Sample Input
610374122
Sample Output
5
一些牛从左到右排列，所有的牛都从左往右看，左边的牛只能看到右边的比它身高严格小的牛的发型，如果被一个大于等于它身高的牛挡住，，那么它就不能看到再右边的牛。要求每头牛可以看到其他牛的总数，转化一下，其实就是求每头牛被看到的总次数。可以用单调栈，每次删除栈中比当前牛的身高小于等于的数。
#include<iostream>#include<stdio.h>#include<stdlib.h>#include<string.h>#include<algorithm>using namespace std;#define maxn 80600int a[maxn],stack[maxn];int main(){int n,m,i,j,top;__int64 sum;while(scanf("%d",&n)!=EOF){memset(stack,0,sizeof(stack));top=0;sum=0;for(i=1;i<=n;i++){scanf("%d",&a[i]);while(top>0 && a[i]>=stack[top])top–;sum+=top;stack[++top]=a[i];}printf("%I64d\n",sum);}return 0;}
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只要笑一笑，没什么事请过不了
